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Qwestrum Engineering360 · Civil Engineering · Steel Structures

Tension Member Design

Compute both the gross-section yield strength and the net-section rupture strength (deducting bolt holes), and take the design capacity as the smaller of the two.
Exam tip: keep SI units consistent end-to-end, write the governing relation symbolically before substituting, and sanity-check magnitude and sign.

Key formulas & points

Skim these first — then read the full notes below.

  • Designstrength=min(Td,Tdb)Design strength = min(T_{d}, T_{db})
  • Staggeredbolts:effectivenetareaAn=A1+(p4g)(A2−A1)Staggered bolts: effective net area A_{n} = A_{1} + (\frac{p}{4g})(A_{2} - A_{1})
  • Slenderness ratio limit: L/r ≤ 400 (recommended)

Topic details

Introduction

Tension member design in IS 800:2007 checks two independent failure modes and adopts the lower capacity. Gross-section yielding governs away from connections, while net-section rupture governs at bolted joints where holes reduce the area.

Scope in B.Tech and GATE syllabus

Gross yield uses the yield stress f_y with material factor γ_m0 = 1.10; rupture uses the ultimate stress f_u with the larger factor γ_m1 = 1.25 because fracture is a brittle, less-forgiving failure. The net area subtracts the bolt-hole area from the gross.

Why this topic matters in practice

When bolts are staggered, the failure path zig-zags, so the net area is corrected by the p²/4g term for each diagonal segment. Block shear is a further check that combines shear and tension along the bolt group perimeter.

Key relations & formulas

Td=Agfyγm0T_{d} = A_{g} \frac{f_{y}}{\gamma_{m0}}
(gross section yield, IS 800)
Tdb=0.9Anfuγm1T_{db} = 0.9 A_{n} \frac{f_{u}}{\gamma_{m1}}
(net section rupture)
An=Ag−Σ(d×t)A_{n} = A_{g} - Σ(d \times t)
(net area; d = bolt hole diameter)

Notation and sign conventions

Relation 1 —
Td=Agfyγm0T_{d} = A_{g} \frac{f_{y}}{\gamma_{m0}}
Td=Agfyγm0T_{d} = A_{g} \frac{f_{y}}{\gamma_{m0}}
(gross section yield, IS 800)
Write this relation with symbols exactly as in Design of Steel Structures — SK Duggal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
Tdb=0.9Anfuγm1T_{db} = 0.9 A_{n} \frac{f_{u}}{\gamma_{m1}}
Tdb=0.9Anfuγm1T_{db} = 0.9 A_{n} \frac{f_{u}}{\gamma_{m1}}
(net section rupture)
Write this relation with symbols exactly as in Design of Steel Structures — SK Duggal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
An=Ag−ΣA_{n} = A_{g} - Σ
An=Ag−Σ(d×t)A_{n} = A_{g} - Σ(d \times t)
(net area; d = bolt hole diameter)
Write this relation with symbols exactly as in Design of Steel Structures — SK Duggal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

A member in pure tension can fail either by yielding over its full cross-section (a ductile, gradual failure) or by rupturing across the reduced net section at bolt holes (a sudden, brittle failure). The code guards against both by computing each capacity and using the minimum.

Governing relations in practice

Gross-section yield strength T_d = A_g f_y / γ_m0 uses the entire cross-section because away from holes the whole area participates. Net-section rupture strength T_db = 0.9 A_n f_u / γ_m1 applies at the connection where the effective area is A_n = A_g − Σ(d·t), with d the hole diameter.

Design and analysis considerations

Staggering bolts spreads the holes so that no single cross-section loses too much area; the p²/4g adjustment credits back some area for the diagonal failure path, allowing more compact and stronger connections.

Advanced theory and extensions

Block shear failure, where a block of material tears out along a combination of shear and tension planes through the bolt group, is a third mode that can govern in short, heavily bolted connections and must be checked separately.

Assumptions and validity limits

State assumptions explicitly before using any relation for tension member design — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Steel Structures viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Steel Structures papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to tension member design.
4. Use equation 1:
Td=Agfyγm0T_{d} = A_{g} \frac{f_{y}}{\gamma_{m0}}
.
5. Use equation 2:
Tdb=0.9Anfuγm1T_{db} = 0.9 A_{n} \frac{f_{u}}{\gamma_{m1}}
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Tension Member Design appears in industrial sheds and high-rise frames. In Indian civil curricula this topic is tested because it connects theory to design of steel members and connections.
GATE and semester exams often combine tension member design with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use tension member design?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Using γ_m0 for the rupture check instead of γ_m1.
• Forgetting to deduct bolt-hole area for the net section.
• Omitting the p²/4g correction for staggered bolt arrangements.
• Neglecting block shear, which can be the lowest capacity in compact connections.

Quick revision checklist

Before attempting tension member design problems, confirm you can:
1.
Designstrength=min(Td,Tdb)Design strength = min(T_{d}, T_{db})

2.
Staggeredbolts:effectivenetareaAn=A1+(p4g)(A2−A1)Staggered bolts: effective net area A_{n} = A_{1} + (\frac{p}{4g})(A_{2} - A_{1})

3. Slenderness ratio limit: L/r ≤ 400 (recommended)
Revise the solved examples in Design of Steel Structures — SK Duggal and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Design tension capacity of a plate

Problem

A 200 × 10 mm plate of Fe410 steel (f_y = 250, f_u = 410 MPa) has one line of 20 mm bolts (hole 22 mm). Find the design tensile strength (gross yield and net rupture) and the governing value.

Solution

Gross area A_g = 200 × 10 = 2000 mm². Gross yield T_d = A_g f_y / γ_m0 = 2000 × 250 / 1.10 = 454 545 N = 454.5 kN. Net area A_n = 2000 − (22 × 10) = 1780 mm². Net rupture T_db = 0.9 A_n f_u / γ_m1 = 0.9 × 1780 × 410 / 1.25 = 525 456 N = 525.5 kN. Design strength = min(454.5, 525.5) = 454.5 kN, so gross-section yielding governs here.

Conceptual check — Tension Member Design

Problem

In a Steel Structures semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of tension member design." What should a complete answer include?

Exams & GATE

IS 800:2007 — check both yield and rupture; governing is lower value.

📖 Standard books (India)

  • Design of Steel Structures — SK Duggal

    Read: Syllabus unit

    IS 800 steel design for Indian practice

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